What is the resultant #pH# of the solution.....?

What is the resultant #pH# of the solution prepared by taking a #10*mL# volume of #HCl(aq)# of #0.10*mol*L^-1# concentration, and diluting this to a #1*L# volume?

1 Answer
anor277
Apr 21, 2017

#pH=3#

Explanation:

#pH=-log_(10)[H_3O^+]#

#[H_3O^+]=(10xx10^-3*Lxx0.1*mol*L^-1)/(1*L)#

#10^-3*mol*L^-1#

And thus #pH=-log_(10)[H_3O^+]#

#-log_(10)(10^-3)=3...........#

And I did this all without a calculator. Note that normally, we would always add acid to water, and not water to acid. However, here the acid concentration is sufficiently dilute to get away with reverse addition.

Related questions
See all questions in pH
Impact of this question
2234 views around the world
You can reuse this answer
Creative Commons License